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# Find the sum of the first 50 terms of the arithmetic sequence if the first term is 21 and the twentieth term is 154.​

• Réponse publiée par: shannel99

9625

Step-by-step explanation:

for an arithmetic sequence

●an = a + (n-1) d

●sn = n/2[2a + (n-1)d]

where is the first term and d the common difference we require to find d

given: a=21 then

a20=21 + 19d = 154

19d = 154 - 21 = 133 d= 133/19 = 7

s50 = 50/2 [(2x21) + (49×7)]

=25(42 + 343)

=25 × 385=9625

• Réponse publiée par: elaineeee

50th term is D. 9625 Explanation:

for an arithmetic sequence

∙xan=a+(n−1)d∙xSn=n2[2a+(n−1)d

]

where a is the first term and d the common difference

we require to find d

given

a=21 then

a20=21+19

d=15419

d=154-21=133

d=13319=7

S50=502[(2×21)+(49×7)]

×=25(42+343)×=25×385=9625

• Réponse publiée par: villatura

b.9765

Step-by-step explanation:

find the sum of the first 50 terms of the arithmetic sequence if the first term is 21 and the twentieth term is 154

9625

Step-by-step explanation:

for an arithmetic sequence

●an = a + (n-1) d

●sn = n/2[2a + (n-1)d]

where is the first term and d the common difference we require to find d

given: a=21 then

a20=21 + 19d = 154

19d = 154 - 21 = 133 d= 133/19 = 7

s50 = 50/2 [(2x21) + (49×7)]

=25(42 + 343)

=25×385=9625

• Réponse publiée par: saintjohn

Hope this helps!

#CarryOnLearning • Réponse publiée par: reyquicoy4321

Step-by-step explanation:

9625

I used other steps. Since first term is 21 and 20th term is 154 its common difference was 7. Using 49 multiplied to 7 and adding 21 you can get the last term 364.By adding last and first and multiply it by 25 you can get the sum of 9625. Hope this helps.

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Find the sum of the first 50 terms of the arithmetic sequence if the first term is 21 and the twenti...